College Algebra: Class 4 Radical Equations Objectives: Solve radical equations Solve equations quadratic in form Solve equations by factoring

Radical Equations and Equations in Quadratic Form Radical Equations – equations that contain a variable under a radical Recall that fractional exponents are radicals. Ex: ½ power is a square root, 1/3 power is a cube root etc. Beware of Extraneous Solutions – extra answers which do not make the equation true. Must check all answers in the original equation.

Process of Solving Radical Equations: Isolate the radical (x + 3) 1/2 – 6 = 3 (x + 3) 1/2 = 9 Raise both sides to a power to cancel the radical ((x + 3) 1/2 ) 2 = 9 2 X + 3 = 81 (square and square root cancel) Solve remaining equation X = 78 Always check your answers. (78 + 3) 1/2 – 6 = 3

Examples: Solve each of the following. (2x – 4) (1/3) – 2 = 0 Isolate the radical: (2x – 4) 1/3 = 2 Raise both sides to the power to eliminate the radical: ((2x-4) 1/3 ) 3 = 2 3 Radical and power cancel: 2x – 4 = 8 Solve the remaining equation: 2x=12 so x=6 Check: (12-4) 1/3 – 2 = 0

(x – 1) (1/2) = x – 7 Radical is already isolated. Square both sides. Remember to square the quantity (x – 1) = (x – 7) 2 x– 1 = x 2 – 14x = x 2 – 15x + 50 Solve the quadratic equation formed 0 = (x – 10)(x – 5) x – 10 = 0 x – 5 = 0 x = 10 x = 5 Check answers (10 – 1) 1/2 = 10 – 7 (5 – 1) 1/2 = 5 – 7 5 does not check, solution is 10

(2x + 3) (1/2) – (x + 2) (1/2) = 2 (2x + 3) 1/2 = 2 + (x + 2) 1/2 (isolate one radical) ((2x + 3) 1/2 ) 2 = (2 + (x + 2) 1/2 ) 2 (square both sides) 2x + 3 = 4 + 4(x + 2) 1/2 + x + 2 (distribute quantity) 2x – x + 3 – 4 – 2 = 4(x + 2) 1/2 (isolate 2 nd radical) X – 3 = 4(x + 2) 1/2 (combine like terms) (X – 3) 2 = (4(x + 2) 1/2 ) 2 (square both sides) X 2 – 6x + 9 = 16(x + 2) (distribute quantity) X 2 – 22x – 23 = 0 (set equal to zero) (x – 23)(x + 1) = 0 (solve quadratic equation) X = 23 x = does not check, 23 is solution

Process for Equations in Quadratic Form Make sure the equation is in standard form. Substitute a variable in for the quadratic value. Solve the quadratic equation formed by the method of your choice. Replace the original value back in for the variable used in the second step. Check

Examples (x + 2) (x + 2) – 12 = 0 U = x + 2 (set u = linear term) U u – 12 = 0 >>>> (u + 12)(u – 1)=0 (substitute in and solve) U = -12 u = 1 X + 2 = -12 x+ 2 = 1 (replace original variable back in and solve) X = -14 x = -1 (check answers – both of these do check)

X 4 – 5x = 0 u = x 2 (set u = linear term) U 2 – 5u + 4 = 0 (u – 4)(x – 1) = 0 (substitute in and solve) U = 4 u = 1 X 2 = 4 x 2 = 1 x = 2, -2, 1, -1 (replace original variable back in and solve)

-7X 3 – 8 = -X 6 x 6 – 7x 3 – 8 = 0 u = x 3 (set u = linear term) u 2 – 7u – 8 = 0 (u – 8)(u + 1) = 0 (solve) u = 8 u = -1 x 3 = 8 x 3 = -1 (replace original variable back in) x = 2 x = -1 (both answers check)

More examples 1/(x + 1) 2 = 1/(x + 1) + 2 u = 1/(x + 1) 1/(x + 1) 2 – 1/(x + 1) – 2 = 0 Standard Form u 2 – u – 2 = 0 Solution for u after substitution u = 2, u = -1 X = -1/2 x = -2 Both answers check

Solve. If you have difficulties please see me for help. X + 2x (1/2) – 3 = 0 What would ‘u’ be substituted in for? (x 1/2 ) Solve then hit return to check your answer X = 1 (9 did not check) 2x -2 – 3x = 0 u = x -1 2u 2 – 3u – 4 = 0 Solve then hit return to check your answer X = (-3 – sqrt 41) / 8 x = (-3 + sqrt 41) / 8

Look over example 8 – Page 122. What would be another way to solve this equation? (POLY) Assn: Page 123 #7, 13, 19, 25, 33, 45, 49, 59, 95